Are $\int_0^t \text{sign}(W_u) \, dW_u$ and $W_t$ independent for a Brownian motion $(W_t)_{t \geq 0}$?

Let $$(W_t)_{t \geq 0}$$ be a Brownian motion. Consider $$X_t = \int^t_0 \text{sign}(W_u) \, dW_u$$ where $$\text{sign}(x) := \begin{cases} 1, & x \geq 0, \\ -1, & x>0. \end{cases}$$ Prove that $$X_t$$ and $$W_t$$ are independent.

How do I prove that? Naturally I need to compute $$E[X_tW_t]=0$$. I don't know if it helps me but I have computed $$X^2_t-t$$ which is a martingale so by Levy characterization Theorem $$X_t$$ is a brownian motion itself.

• $E(X_tW_t)=E\left(\int_0^t\operatorname{sgn}(W_s)dW_s\int_0^tdW_s\right)=E\int_0^t\operatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent. – AddSup Jan 22 at 12:14
• As a consequence of Tanaka's formula, $\int_0^t\text{sign}(W_s)\,dW_s\le |W_t|$ almost surely. This would seem to preclude the independence of these random variables. – John Dawkins Jan 22 at 15:03

It might be tempting to reason as follows: Since $$(X_t)_{t \geq 0}$$ and $$(W_t)_{t \geq 0}$$ are Brownian motions, we know that $$W_t$$ and $$X_t$$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $$\mathbb{E}(X_t W_t) = 0$$, and consequently $$X_t$$ and $$W_t$$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $$X_t$$ and $$W_t$$ are independent.

The reasoning is wrong. Why? In order to conclude that $$W_t$$ and $$X_t$$ are independent, we need to know that the random vector $$(X_t,W_t)$$ is Gaussian; it is not enough to know that (the marginals) $$X_t$$ and $$W_t$$ are Gaussian.

In fact, $$W_t$$ and $$X_t$$ are not independent. By Itô's formula, we have $$W_t^2 = 2 \int_0^t W_s \,dW_s + t.$$ As $$\mathbb{E}(X_t)=0$$ we thus find

\begin{align*} \mathbb{E}(W_t^2 X_t)& =2 \mathbb{E} \left( X_t \int_0^t W_s \, dW_s \right) \\ &=2 \mathbb{E} \left(\left[ \int_0^t \text{sgn}(W_s) \, dW_s \right] \left[ \int_0^t W_s \, dW_s \right] \right) \end{align*}

Applying Itô's isometry we obtain that

$$\mathbb{E}(W_t^2 X_t) =2 \mathbb{E} \left( \int_0^t W_s \, \text{sgn}(W_s) \, ds \right) =2\int_0^t \mathbb{E}(|W_s|) \, ds.$$

The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $$W_s \sim N(0,s)$$ entails $$\mathbb{E}(|W_s|) = \sqrt{(2s)/\pi}$$). As $$\mathbb{E}(X_t)=0$$ this shows that $$\mathbb{E}(W_t^2 X_t) \neq \mathbb{E}(W_t^2) \mathbb{E}(X_t),$$ and therefore $$W_t$$ and $$X_t$$ are not independent.

• Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct? – k.dkhk Jan 23 at 12:34